2 Fixed typos

I am going to stick with the standard terminology $H^m$ here. Taking Fourier transforms one finds that $$\langle u,v\rangle_m=\int\hat u(\xi)\bar{\hat v}(\xi)(1+|\xi|^2)^m\,d\xi$$ (give or take the odd multiplicative constant), where $H^m$ consists precisely of those $u\in L^2$ for which $\langle u,u\rangle<\infty$. This works even for $m<0$, if you allow distributions whose Fourier transforms are functions. Everything follows from this, including the fact that $H^{-m}$ acts as the dual of $H^m$ simply by the distribution $u$ acting on the function $v$, which corresponds to the integral $$\langle u,v\rangle=\int \hat u(\xi)\hat{\bar u(\xi)\bar{\hat v}(\xi)\,d\xi=\int \hat u(\xi)(1+|\xi|^2)^{-m/2}\cdot\hat{\bar v}(\xi)(1+|\xi|^2)^{m/2}\,d\xi$$ where I have split up the integrand into a product of two $L^2$ functions.

For this reason, it seems more natural to identify $H^{-m}$ with the dual of $H^m$ than to identify $H^m$ with its own dual. However, you can go ahead and identify any Sobolev space with the dual of any other just by inserting a suitable power of $1+|\xi|^2$ in the integral defining the pairing between the two.

Rather than coming straight out and answering your question, I'll leave it to you to ponder the consequences of the above. In particular, note that you we embed and identify you have to keep careful track of what space you have identified with whose dual, or you will be endlessly befuddled.

Addendum: To spell out a more direct answer to your question, $\langle\cdot,\cdot\rangle_m$ can identify $H^{m+\sigma}$ with the dual of $H^{m-\sigma}$, since we can write $$\langle u,v\rangle_m=\int \hat u(\xi)(1+|\xi|^2)^{(m-\sigma)/2}\cdot\bar{\hat v}(\xi)(1+|\xi|^2)^{(m+\sigma)/2}\,d\xi$$ where I have split the integrand into a product of two $L^2$ functions.

Edit: Changed a couple $\hat{\bar v}$ into $\bar{\hat v}$.

1

I am going to stick with the standard terminology $H^m$ here. Taking Fourier transforms one finds that $$\langle u,v\rangle_m=\int\hat u(\xi)\bar{\hat v}(\xi)(1+|\xi|^2)^m\,d\xi$$ (give or take the odd multiplicative constant), where $H^m$ consists precisely of those $u\in L^2$ for which $\langle u,u\rangle<\infty$. This works even for $m<0$, if you allow distributions whose Fourier transforms are functions. Everything follows from this, including the fact that $H^{-m}$ acts as the dual of $H^m$ simply by the distribution $u$ acting on the function $v$, which corresponds to the integral $$\langle u,v\rangle=\int \hat u(\xi)\hat{\bar v}(\xi)\,d\xi=\int \hat u(\xi)(1+|\xi|^2)^{-m/2}\cdot\hat{\bar v}(\xi)(1+|\xi|^2)^{m/2}\,d\xi$$ where I have split up the integrand into a product of two $L^2$ functions.

For this reason, it seems more natural to identify $H^{-m}$ with the dual of $H^m$ than to identify $H^m$ with its own dual. However, you can go ahead and identify any Sobolev space with the dual of any other just by inserting a suitable power of $1+|\xi|^2$ in the integral defining the pairing between the two.

Rather than coming straight out and answering your question, I'll leave it to you to ponder the consequences of the above. In particular, note that you we embed and identify you have to keep careful track of what space you have identified with whose dual, or you will be endlessly befuddled.